Let G be a group of order p^m where p is prime number and m is a positive integer. Show that G contains an element of order p.

foass77W 2021-01-27 Answered
Let G be a group of order pm where p is prime number and m is a positive integer. Show that G contains an element of order p.
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Expert Answer

oppturf
Answered 2021-01-28 Author has 94 answers

Let G be a group of order pm, where p is a prime number and m is a positive integer.
Then,
|G|=pm=ppm1=pn,
Where n=pm1
Then, G is a finite group and the prime p divides the order of G, i.e.,
Proving the result by induction on n,
Let n=1
Then, pm1=1 and hence p=1.
Then, |G|=1
Thus, the group has only one element of order 1. The result holds for n=1
Next, suppose every group with pk elements contains an element of order p for every k Consider the class equation of G, |G|=|Z(G)|+aZ(G)[G:C(a)]
If G=Z(G), then G is abelian and hence the result follows.
Suppose GZ(G)
Then, for every a !in Z(G),
C(a)G, and hence |C(a)|<G.
If for some aZ(G),p|C(a)| then, by the induction hypothesis, there is an element a of order pC(a) and hence G.
Otherwise, for every aZ(G),p|C(a)|.
Then, paZ(G)[G:C(a)] and the class equation implies that p|Z(G)|.
Thus, it follows that Z(G) and hence G contains an element of order p.

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